# Math Help - Imaginary, arg problem

1. ## Imaginary, arg problem

Hi I have this problem writting an expression in polar form, I have no issue finding the correct absolute value, but the argument turns out wrong.
The expression is: $\frac{(2+2i)(1+\sqrt{3})}{3i(\sqrt{12}-2i)}$
I wrote all the factors out in polar form, then added the arguments in the denominator and then subracted the arguments in the nominator, what do I do wrong?

2. Originally Posted by Henryt999
Hi I have this problem writting an expression in polar form, I have no issue finding the correct absolute value, but the argument turns out wrong.
The expression is: $\frac{(2+2i)(1+\sqrt{3})}{3i(\sqrt{12}-2i)}$
I wrote all the factors out in polar form, then added the arguments in the denominator and then subracted the arguments in the nominator, what do I do wrong?
Maybe convert it to $a + ib$ form first.

Also, should that numerator be $(2 + 2i)(1 + \sqrt{3})$ or $(2 + 2i)(1 + \sqrt{3}\color{red}i\color{black})$?

3. Originally Posted by Henryt999
Hi I have this problem writting an expression in polar form, I have no issue finding the correct absolute value, but the argument turns out wrong.
The expression is: $\frac{(2+2i)(1+\sqrt{3})}{3i(\sqrt{12}-2i)}$
I wrote all the factors out in polar form, then added the arguments in the denominator and then subracted the arguments in the nominator, what do I do wrong?

Why don't you first simplify the mess? And I'm assuming you meant $1+\sqrt{3}i$ in the numerator:

$\frac{(2+2i)(1+\sqrt{3}i)}{3i(\sqrt{12}-2i)}=\frac{(1+i)(1+\sqrt{3}i)}{3i(\sqrt{3}-i)}$ $=\frac{2e^{\pi i\slash 4}\cdot4e^{\pi i\slash 3}}{3e^{\pi i\slash 2}\cdot 4e^{-\pi i\slash 6}}$

Now simplify the above expression and you're on (note: I wrote down the principal argument in each case. In the second factor in the denominator you can write $4e^{5\pi i\slash 3}$ instead if you want all the arguments to be positive and in $[0,2\pi]$

4. ## Hey thanks!

First of all thanks to both of you for your help.
You are correct there should be an i there where ProveIt said.
Accutually I have to solve this in three different ways.
By simplifying a+bi
by rewritting it in polar form
By using (as tonio did) Eulers approach.
Also this has to be done without a calculator, (that probably means that the argument rational).
Anyway the issue is that I don´t understand what I do wrong with the polar form thing: I just took the arguments in the denominator that gave me:
$\frac{\pi}{4}+\frac{\pi}{3} = \frac{7\pi}{12}$

then the arguments in the nominator: $\frac{\pi}{2}+\frac{\pi}{6} = \frac{2\pi}{3}$

and then substracted that gave me $\frac{7\pi}{12}-\frac{8\pi}{12} = -\frac{\pi}{12}$

but the correct answer is $\frac{\pi}{4}$
not sure what I´m doing wrong here...

5. $\frac{(2+2i)(1+\sqrt{3}i)}{3i(\sqrt{12}-2i)}$

$= \frac{2 + 2\sqrt{3}i + 2i - 2\sqrt{3}}{6 + 3\sqrt{12}i}$

$= \frac{2[1 - \sqrt{3} + (1 + \sqrt{3})i]}{6 + 6\sqrt{3}i}$

$=$
$\frac{2[1 - \sqrt{3} + (1 + \sqrt{3})i]}{6(1 + \sqrt{3}i)}$

$=$
$\frac{1 - \sqrt{3} + (1 + \sqrt{3})i}{3(1 + \sqrt{3}i)}$

$=$
$\frac{[1 - \sqrt{3} + (1 + \sqrt{3})i](1 - \sqrt{3}i)}{3(1 + \sqrt{3}i)(1 - \sqrt{3}i)}$

You should be able to go from here (the bottom will become a rational number).

6. Originally Posted by Henryt999
First of all thanks to both of you for your help.
You are correct there should be an i there where ProveIt said.
Accutually I have to solve this in three different ways.
By simplifying a+bi
by rewritting it in polar form
By using (as tonio did) Eulers approach.
Also this has to be done without a calculator, (that probably means that the argument rational).
Anyway the issue is that I don´t understand what I do wrong with the polar form thing: I just took the arguments in the denominator that gave me:
$\frac{\pi}{4}+\frac{\pi}{3} = \frac{7\pi}{12}$

then the arguments in the nominator: $\frac{\pi}{2}+\frac{\pi}{6} = \frac{2\pi}{3}$

This is already a mess: in a fraction the upper part is called NUMERATOR and the lower one DENOMINATOR. Now, in the numerator you have $z_1=1+i\,,\,\,z_2=1+\sqrt{3}\Longrightarrow \arg z_1=\arctan(1\slash 1)=\pi\slash 4$ $arg(z_2)=\arctan(\sqrt{3})=\pi\slash 3$ , and in the denominator you have $z_3=3i\,,\,\,z_4=\sqrt{3}-i\Longrightarrow \arg(z_3)=\pi\slash 2\,,\,\,\arg(z_4)=-\pi\slash 6$.

Now you do the maths with the powers and you'll get $\pi\slash 4$ , indeed.

Tonio

and then substracted that gave me $\frac{7\pi}{12}-\frac{8\pi}{12} = -\frac{\pi}{12}$

but the correct answer is $\frac{\pi}{4}$
not sure what I´m doing wrong here...
.